Solving Limits of Integrals: Advice & Tips

In summary, the conversation revolved around using integration by parts to solve a limit involving arctan and sin functions. The individual provided their attempted solution and asked for advice on how to solve for n approaching infinity. The expert responded by breaking down the problem and using the Riemann-Lebesgue Lemma to show that the limit is equal to 0.
  • #1
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I tried to use integration by parts.
I took f(x)=arctan(x) => f'(x)= 1/x^2+1
g'(x)=cos(nx) => g(x)= sin(nx)/n
So I get sin(nx)/n * arctan(x) - integral from 0 to 1 from sin(nx)/n(x^2+1)
How to continue ?
I'm always getting stuck with this kind of exercises ( limits of integrals ) because I don't know how to replace n (infinity) in functions and I noticed that I have to use intervals and inequalities to resolve this kind of limits.
Some ideas?
 

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  • #2
I've moved this thread here to our "Calculus" forum since it involves integration.

I would begin by writing:

\(\displaystyle L=\lim_{n\to\infty}\left(\int_0^1 \arctan(x)\cos(nx)\,dx\right)\)

Using IBP in the integral, let's try:

\(\displaystyle u=\arctan(x)\implies du=\frac{1}{x^2+1}\,dx\)

\(\displaystyle dv=\cos(nx)\,dx\implies v=\frac{1}{n}\sin(nx)\)

And so:

\(\displaystyle \int_0^1 \arctan(x)\cos(nx)\,dx=\left[\frac{\arctan(x)\sin(nx)}{n}\right]_0^1-\frac{1}{n}\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx=\frac{\pi\sin(n)}{4n}-\frac{1}{n}\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx\)

Hence:

\(\displaystyle L=-\lim_{n\to\infty}\left(\frac{1}{n}\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx\right)\)

Now, it seems to me considering an increasingly oscillating but bounded sinusoid (which varies from -1 to 1) we can state:

\(\displaystyle \int_0^1 \frac{\sin(nx)}{x^2+1}\,dx\le 2\int_0^1 \frac{1}{x^2+1}\,dx=\frac{\pi}{2}\)

And so:

\(\displaystyle L=-\lim_{n\to\infty}\left(\frac{\pi}{2n}\right)=0\)
 
  • #3
Riemann-Lebesgue lemma:

If $f$ is a real or complex valued measurable function defined on $I$ and Lebesgue integrable over $I$ then:

$\displaystyle \lim_{|\lambda| \to \infty}\int_{I} f(x) \cos({\lambda x})\,\mathrm{d}{x} = \lim_{|\lambda| \to \infty} \int_{I} f(x) \sin({\lambda x})\,\mathrm{d}{x}=0.$​
 
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Related to Solving Limits of Integrals: Advice & Tips

What is a limit of integration?

A limit of integration is the range of values over which an integral is evaluated. It is typically denoted by the letters "a" and "b" and represents the starting and ending points of the integration.

How do I determine the limits of integration?

The limits of integration can be determined by looking at the given function or problem. They can also be found by setting the function equal to the variable of integration and solving for the limits.

What is the importance of finding the correct limits of integration?

The limits of integration are crucial in determining the area under a curve or the volume of a solid. Using the wrong limits can result in an incorrect solution.

What are some common mistakes when solving limits of integrals?

Some common mistakes include forgetting to include the limits of integration, using the wrong function or variable, and not simplifying the integrand before solving.

Are there any tips for solving limits of integrals?

Some tips for solving limits of integrals include carefully reading the problem and identifying the correct limits, using substitution when necessary, and checking your work for errors.

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